A weighing machine inside a stationary lift reads 50 kgf, when a man stands on it. What would happen to the weighing machine reading, if the lift is moving with constant velocity and constant acceleration?
Asked by Topperlearning User | 5th Jun, 2014, 05:00: PM
Apparent weight, W' = m(g + a), where a is the acceleration with which the lift moves upwards.
(1) When the lift moves with constant velocity, a = 0
Therefore, W' = m(g + 0) = mg = 50 kgf (weight remains the same)
(2) When the lift moves with the acceleration a in the upward direction, we have
W' = m (g + a) 50 kgf
Hence, the apparent weight will increase if the lift will move with acceleration a.
Answered by | 5th Jun, 2014, 07:00: PM
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